Figure 4 $\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j,}~0\leq t\leq 2\pi$

Discuss the graph of the vector function \[ \begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}\qquad 0\leq t\leq 2\pi \end{equation*}

Solution The components of $\mathbf{r}=\mathbf{r}(t)$ are the parametric equations $$\begin{equation*} x=x(t) =\cos t\qquad y=y(t) =\sin t\qquad 0\leq t\leq 2\pi \end{equation*}$$

Since $x^{2}+y^{2}=\cos ^{2}t\,{+}\,\sin ^{2}t=1$, the vector function $\mathbf{r=r}(t)$ traces out a circle with its center at the origin and radius $1$. See Figure 4. The curve begins at $\mathbf{r}( 0) =\cos 0\,\mathbf{i}\,{+}\,\sin 0\,\mathbf{j}=\mathbf{i},$ contains the vectors $\mathbf{r} \!\left( \dfrac{\pi }{2}\right) =\cos \dfrac{\pi }{2}\,\mathbf{i}\,{+}\,\sin \dfrac{\pi }{2}\,\mathbf{j}=\mathbf{j},$ $\mathbf{r}(\pi) =\cos \pi \mathbf{i}\,{+}\,\sin \pi \mathbf{j}=-\mathbf{i},$ and $\mathbf{r}\!\left( \dfrac{3\pi }{2}\right) =-\mathbf{j}.$ It ends at $\mathbf{r}( 2\pi )=\mathbf{i},$ so the positive direction of the circle is counterclockwise.